1 litre of coffee at 80oC need to be cooled to 40oC. Assuming that all the ice melts and no heat is lost to the environment. Take the specific heat capacity of coffee to be 4200J/kgK, density of coffee to be 1000kg/m3 and latent heat of fusion of water to be 3.34 x 105 J/kg. You are required to:

a) Determine the heat lost by the coffee when ice is added and coffee cools to 40oC?
Solution:

b) Determine the heat that is required to melt the ice and raise it temperature to 40oC?
=1 litre of coffee 80oC
=Density of coffee is 1000kg/m3
=1 litre of coffee = density of coffee 1kg/m3
That mean, 1000 litre of coffee is equal to 1000kg/m3
Q= me
=1000×3.34 x 105j/kg
=3.34 x 105=0.0000334j/kg
=1000×0.0000334j/kg
=0.0334kj
For that, the heat that is required to melt the ice and raise it temperature to 40oC is 0.0334kj

c) Determine the amount of ice need to be added to 1 litre of the coffee to cools from 80oC to 40oC?
d) If heat loss to the environment is not neglected, what is the relation between the diameter of cup and amount of ice needed to cools 2 litres coffee from 80oC to 40oC? Explain your answer.

c) To determine the amount of ice needed to cool 1 litre of coffee from 80oC to 40oC, we need to calculate the heat lost by the coffee and then convert it to the amount of ice melted.

First, we need to calculate the initial heat of the coffee (Q_initial):
Q_initial = mass x specific heat capacity x change in temperature
Since the volume of coffee is given as 1 litre and the density is 1000kg/m^3, the mass of the coffee is 1kg.
Q_initial = 1kg x 4200J/kgK x (80oC - 40oC) = 168,000J

Next, we need to calculate the heat required to cool the coffee to 40oC (Q_cooling):
Q_cooling = mass x specific heat capacity x change in temperature
Q_cooling = 1kg x 4200J/kgK x (80oC - 40oC) = 168,000J

Since all the ice melts, the heat lost by the coffee is equal to the heat required to melt the ice (Q_melting):
Q_melting = mass of ice x latent heat of fusion
Q_melting = mass of ice x 3.34 x 10^5 J/kg

Setting the heat lost by the coffee equal to the heat required to melt the ice:
Q_initial + Q_cooling = Q_melting

Substituting the values:
168,000J + 168,000J = mass of ice x 3.34 x 10^5 J/kg

Simplifying:
336,000J = mass of ice x 3.34 x 10^5 J/kg

Dividing both sides by 3.34 x 10^5 J/kg:
mass of ice = 336,000J / (3.34 x 10^5 J/kg) = 1.01 kg

Therefore, to cool 1 litre of coffee from 80oC to 40oC, approximately 1.01 kg of ice needs to be added.

d) If heat loss to the environment is not neglected, the amount of ice needed to cool 2 litres of coffee from 80oC to 40oC would depend on the rate of heat loss from the system to the environment. The larger the diameter of the cup, the larger the surface area exposed to the environment, resulting in a faster rate of heat loss. As a result, more ice would be required to cool the coffee in a cup with a larger diameter compared to a cup with a smaller diameter. Therefore, there would be a direct relationship between the diameter of the cup and the amount of ice needed to cool the coffee.

c) To determine the amount of ice needed to cool the coffee from 80oC to 40oC, we need to calculate the heat lost by the coffee when it cools down.

First, let's calculate the heat lost by the coffee using the formula Q = mcΔT, where Q is the heat lost, m is the mass of the coffee, c is the specific heat capacity of coffee, and ΔT is the change in temperature.

Given:
- Volume of coffee = 1 litre = 1 kg (since density of coffee is 1000 kg/m^3)
- Specific heat capacity of coffee = 4200 J/kgK
- Initial temperature of coffee = 80oC
- Final temperature of coffee = 40oC

The change in temperature is:
ΔT = Final temperature - Initial temperature = 40oC - 80oC = -40oC

Let's calculate the heat lost:
Q = mcΔT
Q = 1 kg × 4200 J/kgK × (-40oC)
Q = -168,000 J

Therefore, the coffee loses 168,000 J of heat when it cools from 80oC to 40oC.

d) If heat loss to the environment is not neglected, the relation between the diameter of the cup and the amount of ice needed to cool 2 liters of coffee from 80oC to 40oC is not straightforward.

The amount of heat lost to the environment depends on various factors such as the surface area of the cup, the insulating properties of the cup, the initial and final temperatures of the coffee, and the ambient temperature.

However, in general, a larger diameter cup may result in a larger surface area, facilitating more heat transfer to the environment. This could potentially affect the rate at which the coffee cools and the amount of ice needed.

To accurately determine the relation, you would need to consider the specific heat capacity of the cup, the surface area exposed to the environment, the overall heat transfer coefficient, and other factors related to the specific cup design and heat transfer characteristics.